已知x^2+3x-1=0,那么,代数式x-3/3x^2-6x除以(x+2-5/x-2)的值是多少
问题描述：
已知x^2+3x-1=0,那么,代数式x-3/3x^2-6x除以(x+2-5/x-2)的值是多少
答
x²+3x-1=0
x²+3x=1
于是
(x-3)/(3x^2-6x）/（x+2-5/x-2)
={(x-3)/[3x(x-2)]}/{[(x+2)(x-2)-5]/(x-2)]}
={(x-3)/[3x(x-2)]}/[(x²-9)/(x-2)]
={(x-3)/[3x(x-2)]}/{[(x+3)(x-3)]/(x-2)]}
=1/[3x(x+3)]
=1/[3(x²+3x)]
=1/3